In a resonance tube experiment, resonance occures first at `32.0 cm` and then at `100.0cm`. What is the percentage error in the measurement of end correction.

Options

A. `2.5%`
B. `5%`
C. `7.5%`
D. `10%`

Explanation

To solve the problem of finding the percentage error in the measurement of end correction in a resonance tube experiment, we will follow these steps:

### Step 1: Understand the given data
We have two resonance lengths:
- First resonance length (L1) = 32.0 cm
- Second resonance length (L2) = 100.0 cm

### Step 2: Calculate the difference in lengths
The difference in lengths (ΔL) between the two resonance points is:
\[ \Delta L = L2 - L1 = 100.0 \, \text{cm} - 32.0 \, \text{cm} = 68.0 \, \text{cm} \]

### Step 3: Calculate the average length (L3)
The average length (L3) can be calculated as:
\[ L3 = \frac{L1 + L2}{2} = \frac{32.0 \, \text{cm} + 100.0 \, \text{cm}}{2} = \frac{132.0 \, \text{cm}}{2} = 66.0 \, \text{cm} \]

### Step 4: Calculate the end correction (e)
The end correction (e) is given by the formula:
\[ e = \frac{\Delta L}{2} = \frac{68.0 \, \text{cm}}{2} = 34.0 \, \text{cm} \]

### Step 5: Determine the uncertainty in measurements
Assuming the uncertainty in each measurement (dL1 and dL2) is 0.1 cm, we can express the total uncertainty in the end correction as:
\[ dL = dL2 + 3dL1 = 0.1 \, \text{cm} + 3(0.1 \, \text{cm}) = 0.1 \, \text{cm} + 0.3 \, \text{cm} = 0.4 \, \text{cm} \]

### Step 6: Calculate the percentage error
The percentage error in the end correction is given by:
\[ \text{Percentage Error} = \frac{dL}{2e} \times 100 \]
Substituting the values we have:
\[ \text{Percentage Error} = \frac{0.4 \, \text{cm}}{2 \times 34.0 \, \text{cm}} \times 100 \]
Calculating this gives:
\[ \text{Percentage Error} = \frac{0.4}{68.0} \times 100 = 0.005882 \times 100 \approx 0.5882\% \]

### Conclusion
The percentage error in the measurement of end correction is approximately **0.59%**.

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